Home
Class 12
CHEMISTRY
50 ML of 0.1 M of CH(3)COOH is being tit...

50 ML of 0.1 M of `CH_(3)COOH` is being titrated against 0.1 M `NaOH`. When 25 mL of NaOH has been added , the pH of the solution will be `"___________"xx 10^(-2)` . (Nearest interger)
(Given ,`pK_(a)(CH_(3)COOH)` = 4.76)
log 2 = 0.30
log 3 = 0.48
log 5 = 0.69
log 7 = 0.84
log 11 = 1.04

Text Solution

AI Generated Solution

Promotional Banner

Similar Questions

Explore conceptually related problems

100mL of 0.2M benzoic acid (pk_(a)=4.2) is titrated using 0.2 M NaOH pH after 50mL and 100mL of NaOH have been added are :

100 ml of 0.2 M CH_3COOH is titrated with 0.2 M NaOH solution. The pH of of the solution at equivalent point will be ( pKa of CH_3COOH=4.76 )

The pH of the solution obtained by mixing 250ml,0.2 M CH_3COOH and 200 ml 0.1 M NaOH is (Given pK_a of CH_3COOH = 4.74,log 3=0.48)

100 ml of 0.1 M CH_(3)COOH are mixed with 100ml of 0.1 M NaOH , the pH of the resulting solution would be

100 mL of 0.02M benzoic acid (pK_(a)=4.2) is titrated using 0.02 M NaOH . pH values after 50 mL and 100 mL of NaOH have been added are

The pH of resulting solution when equal volume of 0.01 M NaOH and 0.1 M CH_3COOH are mixed (given pK_a(CH_3COOH)=4.74 and log 3=0.477 ) is